Question

$6.02\times {10}^{22}$ Oxygen Molecules are present at – ${13}^{\circ }C$ in 1500 ml vessel what would be the pressure of the gas?

• A. 4 atm
• B. 2.8 atm
• C. 14.2 atm
• D. 1.42 atm

Answer 1

The correct option is D 1.42 atm

$6.023\times {10}^{23}$ Oxygen Molecules = 1 mole
$6.023\times {10}^{22}$ Oxygen Molecules =$\frac{1\times 6.023\times {10}^{22}}{6.023\times {10}^{23}}$ = 0.1 mole
Hence, no. of moles of Oxygen Molecules in 150ml Vessel = 0.1
We know, PV = n RT or, P = $\frac{nRT}{V}$
$\Rightarrow P=\frac{0.1\times 082\times 260}{1.5}=1.42atm$